# Confused by Paper on the Sum of Three Cubes Problem

I was reading Andrew Broker's paper on his breakthrough with the sum of three cubes problem. In it, I was confused by one result. Earlier in the paper, it is stated that:

$$x^3 + y^3 + z^3 = k \mid k,x,y,z \in \mathbb{Z} \ni k \equiv 3 \textrm{ (mod 9)} \implies x \equiv y \equiv z \equiv 1 \textrm{ (mod 3)}$$

Which totally makes sense to me. However, later, at the very end, it states in the explicit case of ($$k=3$$) with $$|x| \geq |y| \geq |z|$$ without loss of generality, it follows that:

$$z \equiv \frac{4(x+y)^2}{3} + 3((x+y)^2-1) \textrm{ (mod 162)}$$

However, I do not see how $$z \equiv 1 \textrm{ (mod 3)}$$ is ever possible then. For $$\frac{4(x+y)^2}{3}$$ to be integral, 3 must divide $$(x+y)$$, which implies that $$\frac{4(x+y)^2}{3} = 3k$$ for some integer $$k$$. Meanwhile, $$3((x+y)^2-1)$$ is trivially a multiple of 3. Since this is under a modulus which is a multiple of three, wouldn't this imply that $$z$$ must also be a multiple of three, contradicting the first result? They did a computer search for solutions, so I assume I'm missing something, but I don't see what.

There are two small, previously found solutions, (1,1,1 & -5,4,4), but for neither part does the second claim hold true. I would presume it is only for larger solutions, or perhaps solutions where $$x > y > z$$, but at this point I don't know exactly what's going on.

So, in summary, why did they search for primes if $$z$$ cannot be 1 mod 3? Did I misread something or what?

• As for the solutions $(1,1,1)$ and $(-5,4,4)$, they assume that $y\ne z$ in the beginning of the second part. – Milten Mar 28 at 11:20
• You misunderstood the congruence as it is $4(\frac{x+y}{3})(x+y)$ and the first term is a reciprocity symbol, not a division, so $x+y$ actually must be coprime with $3$ as assumed – Conrad Mar 28 at 12:09
• apologies, what is a reciprocity symbol? – Zachary Hunter Mar 28 at 13:35
• For an odd prime $p$ and any integer $a$, $(a/p)=0$ if $p$ divides $a$. If not: $(a/p)=1$ if there are solutions to $x^2\equiv a \pmod p$, else it equals $-1$. i.e. whether you can take squareroot modulo $p$. Also see Legendre symbol – Yong Hao Ng Mar 28 at 13:50